Exercise 1. Eight identical tiles, numbered 1 through 8, are placed in a bag. Th
ID: 2924418 • Letter: E
Question
Exercise 1. Eight identical tiles, numbered 1 through 8, are placed in a bag. The bag is shaken, and three are removed at random. If you treat selections as sets of three, then the size of the sample space is under 100. (A) What is the size of the sample space? (B) What is the chance that the selected numbers sum to 13? (C) What is the chance that the 7 tile is not selected? (D) What is the chance that 7 is the highest value on the selected tiles? (E) What is the chance that no two of the selected tiles have adjacent number? (For example, if the 1 is selected, then the 2 is not, or if the 4 is selected, than neither the 3 nor the 5 is on another selection.) Exercise 2. A bowl contains 7 red marbles, 6 white marbles and 4 blue marbles. From the bowl, five are selected at random. For the selection, assume all marbles are equally likely to be chosen. For each of the following, you may the answer using formulas with combinations, or as fractions. (A) What is the chance that no red marble is selected? (B) What is the chance that exactly two red marbles are selected? (C) What is the chance that only red and white marbles are selected? (D) What is the chance that the selection consists of two red marbles, two blue marbles and one white marble? (E) What is the chance that at least one blue marble is selected?Explanation / Answer
1)
A) here size of the sample space =8C3 =8!/(5!*3!) =56
B)number of ways 13 can come as a sum ={(1,5,7) ,(1,4,8),(2,3,8),(2,4,7),(2,5,6),(3,4,6)} =6
hence probability =6/56 =3/28
C)probability that 7 is not selected =7C3/8C3 =35/56 =5/8
d)probability =7 is selected and 8 is not selected =selecting rest of 2 from 6 numbers below 7 =6C2/8C3 =15/56
e)number of ways ={(1,3,5),(1,4,6),(1,5,7),(1,6,8),(2,4,6),(2,5,7),(2,6,8),(3,5,7),(3,6,8),(4,6,8)} =10
hence probability =10/56 =5/28
2)a) probability of no red marble =10C5 /17C5 =252/6188
b)probability of 2 red marble =10C3 *7C2/17C5 =120*21/6188=2520/6188
c)probability of no blue marble=13C5 /17C5 = 1287/6188
d)probability =6C1 *7C2*4C2/17C5 =6*21*6/6188=756/6188
e)probability of at least one blue mareble =1-P(all red and white marble as in part C)=1-1287/6188=4901/6188
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